1. Key Concept: Binomial Expansion and the Alternating Sum of Coefficients

The binomial expansion of $(1-x)^n$ is given by the formula: $(1-x)^n = \sum_{r=0}^{n} \binom{n}{r} (1)^{n-r} (-x)^r = \sum_{r=0}^{n} \binom{n}{r} (-1)^r x^r$ For this problem, $n=100$. So, the expansion of $(1-x)^{100}$ is: $(1-x)^{100} = \binom{100}{0}x^0 - \binom{100}{1}x^1 + \binom{100}{2}x^2 - \ldots + (-1)^{100}\binom{100}{100}x^{100}$ The coefficient of the $(r+1)$-th term in this expansion is $\binom{100}{r}(-1)^r$.

A fundamental property of binomial coefficients is that the alternating sum of all coefficients in the expansion of $(1-x)^n$ is zero (for $n > 0$): $\sum_{r=0}^{n} (-1)^r \binom{n}{r} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \ldots + (-1)^n \binom{n}{n} = 0$ This can be easily verified by substituting $x=1$ into the original expansion: $(1-1)^n = 0^n = 0$.

2. Understanding the Problem: Sum of Coefficients of the First 50 Terms

We need to find the sum of the coefficients of the first 50 terms. The first term corresponds to $r=0$, the second term to $r=1$, and so on, up to the 50th term, which corresponds to $r=49$. Therefore, the sum we are looking for, let's call it $S$, is: $S = \sum_{r=0}^{49} (-1)^r \binom{100}{r} = \binom{100}{0} - \binom{100}{1} + \binom{100}{2} - \ldots - \binom{100}{49}$ (Note: $(-1)^{49} = -1$)

3. Applying the Alternating Sum Property to the Full Expansion

Let's write out the complete alternating sum of coefficients for $(1-x)^{100}$: $\binom{100}{0} - \binom{100}{1} + \binom{100}{2} - \ldots - \binom{100}{49} + \binom{100}{50} - \binom{100}{51} + \ldots + \binom{100}{100} = 0$ We can strategically split this sum into three parts to isolate $S$: $\underbrace{\left( \binom{100}{0} - \binom{100}{1} + \ldots - \binom{100}{49} \right)}_{\text{Sum } S} + \binom{100}{50} + \underbrace{\left( - \binom{100}{51} + \binom{100}{52} - \ldots + \binom{100}{100} \right)}_{\text{Let this be } S' } = 0$ So, we have the equation: $S + \binom{100}{50} + S' = 0$.

4. Utilizing the Symmetry Property of Binomial Coefficients

The symmetry property states that $\binom{n}{r} = \binom{n}{n-r}$. We will use this to relate $S'$ to $S$. For $n=100$:

• $\binom{100}{51} = \binom{100}{100-51} = \binom{100}{49}$
• $\binom{100}{52} = \binom{100}{100-52} = \binom{100}{48}$
• ...
• $\binom{100}{99} = \binom{100}{100-99} = \binom{100}{1}$
• $\binom{100}{100} = \binom{100}{100-100} = \binom{100}{0}$

Now, let's rewrite $S'$ by substituting these symmetric equivalents: $S' = - \binom{100}{51} + \binom{100}{52} - \ldots - \binom{100}{99} + \binom{100}{100}$ $S' = - \binom{100}{49} + \binom{100}{48} - \ldots - \binom{100}{1} + \binom{100}{0}$ Rearranging the terms in $S'$ to match the order of $S$: $S' = \binom{100}{0} - \binom{100}{1} + \ldots + \binom{100}{48} - \binom{100}{49}$ By careful comparison, we can see that $S'$ is identical to $S$. Therefore, $S' = S$.

5. Solving for the Required Sum

Substitute $S'$ with $S$ back into our main equation from Step 3: $S + \binom{100}{50} + S = 0$ $2S + \binom{100}{50} = 0$ $2S = - \binom{100}{50}$ $S = - \frac{1}{2} \binom{100}{50}$

6. Final Simplification using Combinatorial Identity

The options are given in terms of $\binom{99}{49}$ or $\binom{101}{50}$. We need to convert our result, $S = - \frac{1}{2} \binom{100}{50}$, into one of these forms. A useful combinatorial identity states that for any even $n$, $\binom{n}{n/2} = 2 \binom{n-1}{(n/2)-1}$. In our case, $n=100$, so $n/2 = 50$. Applying the identity: $\binom{100}{50} = 2 \binom{100-1}{50-1} = 2 \binom{99}{49}$ Now, substitute this expression for $\binom{100}{50}$ back into the equation for $S$: $S = - \frac{1}{2} \left( 2 \binom{99}{49} \right)$ $S = - \binom{99}{49}$

7. Tips for Success and Common Mistakes

• Mind the Signs: When dealing with alternating sums, meticulously track the sign of each term. A single sign error can lead to an incorrect answer.
• Identify the Correct Range: "First 50 terms" refers to $r=0$ through $r=49$ in the general term $(-1)^r \binom{n}{r}x^r$.
• Leverage Properties: Always look for opportunities to use fundamental properties like the alternating sum identity ($\sum (-1)^r \binom{n}{r} = 0$) and the symmetry property ($\binom{n}{r} = \binom{n}{n-r}$). These are powerful tools for simplifying complex sums.
• Central Term: For even $n$, the term $\binom{n}{n/2}$ (the middle term) is unique in that it does not have a symmetric counterpart with an opposing sign that would allow for direct cancellation within a perfectly balanced alternating sum. It often needs to be handled separately, as demonstrated in this solution.
• Combinatorial Identities: Be familiar with common combinatorial identities, especially those involving adjacent or central binomial coefficients. Pascal's identity ($\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$) and its variations are frequently used.

8. Summary/Key Takeaway

This problem is a classic example of how to combine the alternating sum property and the symmetry property of binomial coefficients to solve for a partial sum. By recognizing that the sum of the first half of the alternating terms (excluding the middle term) is the negative of the sum of the second half, we can quickly deduce the value of the desired sum. The final step involves applying a standard combinatorial identity to match the result with the provided options.

The final answer is $\boxed{{-}^{99} \mathrm{C}_{49}}$.